Limit of Integration (original) (raw)
Last Updated : 1 Nov, 2025
Integration is a mathematical concept used to find the accumulation or total of a quantity, often represented by a function, over a specified interval. It involves finding the antiderivative (or indefinite integral) of a function.
In simpler terms, integration helps calculate the area under a curve on a graph or the net change in a quantity. Limits of integration are the numbers that set the boundaries for calculating the definite integral of a function. The definite integral, ∫f(x)dx, involves finding the antiderivative F(x) and then evaluating it at the upper and lower limits, [a, b].
The limits of integration are the numbers we use to set the range for integrating a function. When we integrate a function, which is essentially the opposite of differentiation, we obtain what's called an antiderivative.To figure out the definite integral between two points, say [a, b], we subtract the antiderivative's value at (b) from its value at (a). In this range, (a) is the upper limit, and (b) is the lower limit.
\int_{a}^{b} f(x) \,dx = F(b) - F(a)
This process helps us find the area under the curve between two points. It's like calculating the space enclosed by the curve.
Upper Limits of Integration
The upper limit of integration refers to the higher endpoint in a specified range when calculating a definite integral. In the context of integration, particularly in ∫ab f(x) dx, the upper limit is represented by 'b'.
This signifies the value at which the integration process concludes. When evaluating the integral between 'a' and 'b', we find the antiderivative of the function and subtract the value of this antiderivative at 'a' from its value at 'b'.
Lower Limits of Integration
The lower limit of integration is one of the numbers that defines the range for calculating the definite integral of a function. When we perform definite integration ∫ab f(x) dx, where a and b are limits, we find the antiderivative F(x) and then evaluate it at the upper limit (F(b)) and subtract the value at the lower limit (F(a)).
In this context, the number 'a' is referred to as the lower limit, marking the starting point of the interval, while 'b' is the upper limit, representing the endpoint.
Steps to Find the Limits of Integration
To find the limits of integration, we can use the following steps for any integral.
First, we solve the integration problem by figuring out the antiderivative of a function, represented as **∫ a b f(x).dx = [F(x)] a b.
The second step is applying the limits [a, b] to the antiderivative.
- This means substituting the values of a and b into the antiderivative.
- The final result is obtained by subtracting F(a) from F(b), expressed as [F(x)] evaluated from a to b i.e., F(a) − F(b).
In simple terms, the limits of integration help us find the specific numerical value of the given integral expression.
How to Find Upper and Lower Limit of Integration
If you are given a definite integral like ∫ab f(x) dx, where f(x) is the function and a and b are the limits of integration, then:
- **a is the lower limit of the integration, and
- **b is the upper limit of the integration.
How to Change the Limits of Integration?
Changing the limits of integration is a process that involves a few simple steps:
- **Identify Original Limits: Pinpoint the initial integration limits, commonly labeled as 'a' and 'b.' These values indicate the starting and ending points of integration.
- **Determine New Limits: Choose new limits, typically represented as 'c' and 'd.' 'c' becomes the updated lower limit, while 'd' becomes the new upper limit.
- **Adjust the Integral: Subtract 'a' from both upper and lower limits, and also subtract 'c' from both. This adjustment ensures the integral remains consistent.
- **Add the Difference to the Integral: The subtraction in the previous step yields a difference, referred to as 'k.' Add 'k' to the integral to accommodate the adjusted limits.
- **Substitute the New Limits: Replace the original limits with the new values. Substitute 'a' with 'c' and 'b' with 'd' in the integral expression.
Formulas of Limits of Integration
Following are the formulas for the limits of integration:
1. \int_{a}^{b} f(x) \,dx = \int_{a}^{b} f(t) \,dt
- This says that integrating a function from (a) to (b) with respect to (x) is the same as integrating the same function from (a) to (b) with respect to (t).
2. \int_{a}^{b} f(x) \,dx = -\int_{b}^{a} f(x) \,dx
- This formula means that changing the limits of integration also changes the sign of the integral.
3. \int_{a}^{b} c \cdot f(x) \,dx = c \cdot \int_{a}^{b} f(x) \,dx
- If you have a constant (c) multiplied by the function, you can take the constant outside the integral.
4. \int_{a}^{b} [f(x) \pm g(x)] \,dx = \int_{a}^{b} f(x) \,dx \pm \int_{a}^{b} g(x) \,dx
- This formula allows you to split the integral when you have the sum or difference of two functions.
5. \int_{b}^{a} f(x) \,dx = \int_{a}^{b} f(a + b - x) \,dx
- Integrating from (b) to (a) is the same as integrating from (a) to (b) with a change in the variable.
6. \int_{a}^{0} f(x) \,dx = \int_{a}^{0} f(a - x) \,dx
- This formula involves integrating from (a) to 0, and it's related to the one above.
7. \int_{2a}^{0} f(x) \,dx = 2 \cdot \int_{a}^{0} f(x) \,dx
- If f(2a - x) = f(x), then the integral from (2a) to 0 is twice the integral from (a) to 0.
8. \int_{2a}^{0} f(x) \,dx = 0
- If f(2a - x) = -f(x), then the integral from (2a) to 0 is zero.
9. \int_{-a}^{a} f(x) \,dx = 2 \cdot \int_{0}^{a} f(x) \,dx
- If f(x) is an even function f(-x) = f(x), the integral from (-a) to (a) is twice the integral from (0) to (a.
10. \int_{-a}^{a} f(x) \,dx = 0
- If f(x)is an odd function f(-x) = -f(x), the integral from (-a) to (a) is zero.
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Solved Examples of Limits of Integration
**Example 1: You have the function f(x) = 2x + 1, and you want to find the area under the curve between the points x = 1 and x = 3. Write and solve the definite integral for this scenario.
**Solution:
To solve : I = \quad \int_{1}^{3} (2x + 1) \,dx
First, find the antiderivative of (2x+1), which is (x2+x).
⇒I = \left[x^2 + x\right]_{1}^{3}
Now substitute the upper limit x=3 and lower limit x=1:
⇒ I = [32 + 3] - [ 12 + 1]
⇒ I = [9+3] - [1+1]
⇒ I = 12-2
⇒ I = 10∴ the area under the curve of 2x + 1 from x=1 to x=3 is 10 square units.
**Example 2: Consider an object's velocity given by v(t) = 3t2 where (t) is time. Find the distance traveled by the object from t=1 to t=2. Express the solution as a definite integral.
**Solution:
Let I = \quad \int_{1}^{2} 3t^2 \,dt
First, find the antiderivative of 3t2, which is t3.
⇒ I = \left[t^3\right]_{1}^{2}
Now substitute the upper limit t=2 and lower limit t=1:
⇒ I = (2)3 - (1)3
⇒ I = 8-1
⇒ I = 7∴ the distance traveled by the object from (t=1) to (t=2) is 7 units.
Unsolved Questions on Limits of Integration
**Question 1: Evaluate the integral using the limits of integration: I = \int_{0}^{2} (3x^2 + 2x + 1) \, dx
**Question 2 : The velocity of a car at time t seconds is given by v(t) = 4t + 2m/s. Find the total distance covered by the car from t = 0 to t = 5 seconds.Distance = \int_{0}^{5} (4t + 2)dt.
**Question 3: Water is flowing into a tank at a rate of R(t) = 3t2+2t liters per minute. Find the total amount of water that flows into the tank between t = 1 minute and t = 3 minutes. Total water \int_{1}^{3} (3t^2 + 2t)dt.
**Question 4: The height of a curve is given by y = x2+ 1 . Find the area under this curve between x = 0and x = 2. Area = \int_{0}^{2} (x^2 + 1) dx.